Download A ball is projected vertically upwards with an initial speed of 10 m s

A ball is projected vertically upwards with an initial speed of 10 m s^-1
at a height of 2.5m above the ground. In part (a) ignore all frictional forces.
(a)
(i) Draw a force diagram for the ball while it is in motion. [1]
When ball is thrown upwards, the weight acts downwards and ball moves up against this force as shown
below.
(ii) Define appropriate coordinate axes and an origin, and state the
initial velocity and initial displacement in terms of the unit
vectors and origin that you have chosen. [2]
Initial velocity = 10m/sec in upward direction= 10az
Initial displacement = 2.5 from ground=(0,0,2.5)
(iii) Determine, in terms of the magnitude of the acceleration due to
gravity, g, the maximum height that the ball reaches above the
point of projection, and the time taken to reach this position. [3]
At maximum height, velocity =0.
Putting in v2=u2-2gh, we get 0=100-2gh
Or, h = 50/g
Thus, ball will reach a height of (50/g) m from the point of projection.
Putting v = u-gt, we get t = 10/g; hence, it will take 10/g secs to reach the maximum height.
(iv) Determine the speed at which the ball hits the ground, correct to
two decimal places, taking the value of g to be 9.81ms^-2.
After reaching the top most point, the ball will start moving down from the height of [(50/g)+2.5] from
ground.
Putting in v2=u2+2gh and putting g = 9.81, we get v = sqrt(100+49.05)=12.208, 12.21m/sec.
Hence, the ball will strike ground with a velocity of 12.21m/sec.
In the remainder of the question revise this model by taking air resistance
into account. Model the ball as a sphere of diameter D and mass m, and
assume that the quadratic model of air resistance applies.
(b) In this part of the question the upward motion of the ball is
investigated. The ball is projected at time t = 0.
(i) Draw a force diagram showing all the forces acting on the ball,
and express each force in terms of the unit vectors using the same
axes as in part (a), justifying your derivation. [2]
Air frictional force (Fr) will be proportional to v2 and it will act in opposite direction to v.
Fr = -kv2, where k is proportionality constant. We can represent the forces as given below.
(ii) Show that the component of acceleration at time t in the upward
direction is given by
a = g/b^2(v^2+ b^2),
where v is the speed of the ball at time t, and b^2 = mg/0.2D^2.
Equation of motion will be given by
m
dv
 mg  kv 2
dt
dv
k
  g  v2
dt
m
dv
g 2
v2
2
  2 v  b    g  g 2
We are given that a 
dt
b
b
Comparing, we get
k
g
mg mg  0.2D2
 2 k  2 
 0.2 D2
m b
b
mg
(iii) By writing a = dv/dt, solve the resultant differential equation and
determine the time t in terms of v, b, g and v0, where v0 is the
initial speed, upwards, of the ball. [4]
dv
g
dv
g
  2 v 2  b 2   2
  2 dt
2
dt
b
v b
b
1
v
g
tan 1   2 t  C
b
b
b
as 
dx
1
x
 tan 1
2
a x
a
a
2
Putting initial condition v = v0 at t =0, we get
Hence,
v
1
tan 1 0  C .
b
b
v
v 1
1
v
g
1
g
1
v
tan 1   2 t  tan 1 0  2 t  tan 1 0  tan 1
b
b
b
b
b
b
b
b b
b
Hence, t 
b(v0  v)
b
tan 1
g
b2  v0v
(iv) By writing a = v dv/dx, solve the resulting differential equation
and determine the height above the point of projection at time t
in terms of v, b, g and v0. [4]
v
dv
g
vdv
g
  2 v 2  b 2   2
  2 dx
2
dx
b
v b
b
Or,
1
g
ln(v 2  b 2 )   2 x  C
2
b
Putting initial condition v = v0 at x =0, we get
1
ln(v0 2  b 2 )  C
2
1
g
1
b2 v0 2  b2
2
2
2
2
ln(v  b )   2 x  ln(v0  b )  x 
ln
2
b
2
2 g v 2  b2
(d) In this part of the question the downward motion of the ball is
investigated. The origin or coordinate system may be changed, in
which case they must be clearly defined.
(i) Draw a force diagram showing all the forces acting on the ball,
and express each force in terms of the unit vectors, justifying your
derivation. [2]
As ball is moving downwards, v and mg will be acting in the same direction. Fr will act upwards.
(ii) Derive the equation of motion (i.e. a differential equation) of the
ball, as it moves downwards, in terms of v, b and g. Hence
calculate the terminal speed of the ball if it could continue to fall
indefinitely beyond the original point of projection. [3]
Equation of motion will be given by
m
dv
 mg  kv 2
dt
 v2 
dv
k
mg
 g  v 2  g  2 v 2  g 1  2 
dt
m
mb
 b 
Or,
dv
g
  2 dt
2
v b
b
Or,
1 v b
g
ln
  2 t C
2b v  b
b
2
as 
1
1
xa
dx 
ln
2
x a
2a x  a
2
Putting initial condition v = v0 at t =0, we get
Hence,
1 v0  b
ln
C
2b v0  b
2g
 t
(v  b)(v0  b)
1 v b
g
1 v b
ln
  2 t  ln 0

e b .
2b v  b
b
2b v0  b
(v  b)(v0  b)
Putting t = ∞, we get v = b, that is, terminal velocity of the ball.
(iii) For the coordinate system that you have chosen, what is the
condition to determine when the ball hits the ground? [1]
v
 v2 
dv
v
g
 g 1  2   2 2 dv   2 dx
dx
b
 b  v b
Or,
1
g
ln(v 2  b 2 )   2 x  C
2
b
Putting initial condition v = v0 at x =H(height from which ball starts falling), we get
1
g
1
g
ln(v0 2  b 2 )   2 H  C  ln(v0 2  b 2 )  2 H .
2
b
2
b
Hence,
1
g
1
g
ln(v 2  b 2 )   2 x  ln(v0 2  b 2 )  2 H
2
b
2
b
When ball strikes ground, x = 0. Hence,
H
1
1
g
v2  b2 2 g
v2  b2
2
2
2
2
b2
ln(v  b )  ln(v0  b )  2 H  ln 2

H


e
.
2
2
b
v0  b 2 b 2
v0 2  b 2
2g